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This happens because now the slit behaves as multiple small slits, and you easily figure it out, see this (let's say the slit as AB and where ever you form fringe be X):

For central maxima: all slits will be in nearly same phase, as all need to travel same nearly distance, so the add-up giving brightest maxima. (AX = BX) img1

For first dark fringe: there will some X on screen such that half part of light (say AO for some O in AB) is in positive phase, and corresponding to each phase there is equally negative phased wave (from OB).img 2

The important part, for next bright fringe: let's divide AB into AM + MN + NB for some M, N in AB. Now the case is that AM and MN cancels each other, but NB is left, giving a bright fringe but lesser bright.img 3

Further, you can divide it into 4 sections for next dark fringe, and 5 sections for bright one (obviously less bright).

i hope it helped :)

This happens because now the slit behaves as multiple small slits, and you easily figure it out, see this (let's say the slit as AB and where ever you form fringe be X):

For central maxima: all slits will be in nearly same phase, as all need to travel same nearly distance, so the add-up giving brightest maxima. (AX = BX)

For first dark fringe: there will some X on screen such that half part of light (say AO for some O in AB) is in positive phase, and corresponding to each phase there is equally negative phased wave (from OB).

The important part, for next bright fringe: let's divide AB into AM + MN + NB for some M, N in AB. Now the case is that AM and MN cancels each other, but NB is left, giving a bright fringe but lesser bright.

Further, you can divide it into 4 sections for next dark fringe, and 5 sections for bright one (obviously less bright).

i hope it helped :)

This happens because now the slit behaves as multiple small slits, and you easily figure it out, see this (let's say the slit as AB and where ever you form fringe be X):

For central maxima: all slits will be in nearly same phase, as all need to travel same nearly distance, so the add-up giving brightest maxima. (AX = BX) img1

For first dark fringe: there will some X on screen such that half part of light (say AO for some O in AB) is in positive phase, and corresponding to each phase there is equally negative phased wave (from OB).img 2

The important part, for next bright fringe: let's divide AB into AM + MN + NB for some M, N in AB. Now the case is that AM and MN cancels each other, but NB is left, giving a bright fringe but lesser bright.img 3

Further, you can divide it into 4 sections for next dark fringe, and 5 sections for bright one (obviously less bright).

i hope it helped :)

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This happens because now the slit behaves as multiple small slits, and you easily figure it out, see this (let's say the slit as AB and where ever you form fringe be X):

For central maxima: all slits will be in nearly same phase, as all need to travel same nearly distance, so the add-up giving brightest maxima. (AX = BX)

For first dark fringe: there will some X on screen such that half part of light (say AO for some O in AB) is in positive phase, and corresponding to each phase there is equally negative phased wave (from OB).

The important part, for next bright fringe: let's divide AB into AM + MN + NB for some M, N in AB. Now the case is that AM and MN cancels each other, but NB is left, giving a bright fringe but lesser bright.

Further, you can divide it into 4 sections for next dark fringe, and 5 sections for bright one (obviously less bright).

i hope it helped :)